---
tag: KJ2984R97Y
kind: definition
name: Volterra equation of convolution type
msc: [45D05, "45E10"]
defines: SNFYTC1TTA
uses: [DND8CBEVXT, B5AB451TB7, 2Z29JN3VES]
sources:
  - url: https://en.wikipedia.org/wiki/Volterra_integral_equation
---

## Definition

Let $T > 0$ or $T = \infty$, let $k \colon (0, T) \to \C$ be locally integrable, let $f \colon [0, T) \to \C$ be a given function, and let $\lambda \in \C$.
A *Volterra equation of convolution type* is a [[Volterra integral equation|DND8CBEVXT]] whose kernel is a [[difference kernel|B5AB451TB7]] $K(t, s) = k(t - s)$.
In the second kind it reads
$$x(t) = f(t) + \lambda \int_0^t k(t - s) \, x(s) \, ds, \quad 0 \le t < T,$$
and in the first kind it reads
$$\int_0^t k(t - s) \, x(s) \, ds = f(t), \quad 0 \le t < T.$$

## Remarks

The integral is the [[convolution|2Z29JN3VES]] $(k * x)(t) = \int_0^t k(t - s) \, x(s) \, ds$ of functions on $[0, \infty)$, so the second kind is written $x = f + \lambda \, k * x$.
Example: $x(t) = 1 + \int_0^t e^{-(t - s)} x(s) \, ds$, with $k(u) = e^{-u}$.
The function $x(t) = 1 + t$ solves it: $\int_0^t e^{-(t - s)} (1 + s) \, ds = t$, by the substitution $u = t - s$ and $\int_0^t e^{-u}(1 + t - u) \, du = t$.
Non-example: $x(t) = 1 + \int_0^t t s \, x(s) \, ds$ has kernel $K(t, s) = ts$, which is not a function of $t - s$ alone.
The [[Laplace transform method|EHGECREYB4]] solves convolution equations on $[0, \infty)$ when $k$ and $f$ have Laplace transforms.